The de-Broglie equation is:

The kinetic energy of an electron is $4.55 \times 10^{-25} \ J$. Find its wavelength.

If the kinetic energy of an electron of mass $9.0 \times 10^{-31} \ kg$ is $2.0 \times 10^{-25} \ J$,the wavelength of the electron in $nm$ is approximately (in $.3$)

The wavelength of a photon is $5200 \ \mathring{A}$. At what velocity will an electron have the same wavelength as this photon (in $m/s$)?

The de Broglie wavelength of an electron with kinetic energy of $2.5 \ eV$ is (in $m$):
$(1 \ eV = 1.6 \times 10^{-19} \ J, m_{e} = 9 \times 10^{-31} \ kg)$

If $a_0$ is the Bohr radius,then the de-Broglie wavelength of an electron revolving in the second excited state of the $H$ atom will be: